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Alecsey [184]
3 years ago
15

Ammonium perchlorate nh4clo4 is a powerful solid rocket fuel, used in the space shuttle boosters. it decomposes into nitrogen n2

gas, chlorine cl2 gas, oxygen o2 gas and water vapor, releasing a great deal of energy. calculate the moles of ammonium perchlorate needed to produce 1.4 of nitrogen. be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Chemistry
1 answer:
Alchen [17]3 years ago
4 0

Answer:

Explanation:

2 NH₄ClO₄ ⇒ N₂ + Cl₂ + 2O₂ + 4 H₂O

2 moles          1 mole

We are required to produce 1.4 moles of N₂ from ammonium perchlorate .  

1 mole of N₂ is produced by 2 moles of ammonium perchlorate

1.4 moles of N₂ will be produced by 2 x 1.4 = 2.8 moles of ammonium perchlorate .

Answer is 2.8 moles .

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Calcium Chloride is an inorganic compound, a salt that can dissolve in water.

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In the diagram, which letter represents the energy of the products?
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Which model of the atom is the most accurate
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4. Magnesium and oxygen undergo a chemical reaction to
erastovalidia [21]

Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

  1. Convert grams of MgO to moles of MgO.
  2. Moles of MgO to moles of O₂
  3. And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

3 0
2 years ago
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